HDU 2682 Tree 最小生成树 prim模板

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=2682

题意：

有n个城市，每个城市有一个幸福值，对任意两个城市，如果两个城市中任意一个城市幸福值为素数或者两者之和为素数，那么这两个城市是可以连接的，代价是 min(min(vi,vj),|vi−vj|) ，求把所有城市连接再一起的最小花费，不能连通输出-1

思路：

最小生成树，测试prim模板 朴素的prim:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <map> using namespace std; typedef pair<int, int> pii; const int N = 600 + 10, M = 2000000 + 10, INF = 0x3f3f3f3f; int g[N][N]; int mincost[N]; bool vis[N], prime[M]; int prim(int n) { memset(mincost, 0x3f, sizeof mincost); memset(vis, 0, sizeof vis); mincost[1] = 0; int ans = 0, num = 0; while(true) { int v = -1; for(int i = 1; i <= n; i++) if(! vis[i] && (v == -1 || mincost[i] < mincost[v])) v = i; if(v == -1 || mincost[v] == INF) break; vis[v] = true; ans += mincost[v]; num++; for(int i = 1; i <= n; i++) mincost[i] = min(mincost[i], g[v][i]); } return num == n ? ans : -1; } void table() { prime[0] = prime[1] = 1; for(int i = 2; i * i < M; i++) if(! prime[i]) for(int j = i + i; j < M; j += i) prime[j] = 1; } int main() { table(); int t, n, a[N]; scanf("%d", &t); while(t--) { memset(g, 0x3f, sizeof g); scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) { if(! prime[a[i]] || ! prime[a[j]] || ! prime[a[i]+a[j]]) { int cost = min(min(a[i], a[j]), abs(a[i] - a[j])); g[i][j] = g[j][i] = cost; } } printf("%d\n", prim(n)); } return 0; }

二叉堆优化的prim:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <map> using namespace std; typedef pair<int, int> pii; const int N = 600 + 10, M = 2000000 + 10, INF = 0x3f3f3f3f; struct edge { int to, cost, next; }g[N*N*2]; int cnt, head[N]; int mincost[N]; bool vis[N], prime[M]; void add_edge(int v, int u, int cost) { g[cnt].to = u, g[cnt].cost = cost, g[cnt].next = head[v], head[v] = cnt++; } int prim(int n) { priority_queue<pii, vector<pii>, greater<pii> > que; memset(vis, 0, sizeof vis); memset(mincost, 0x3f, sizeof mincost); que.push(pii(0, 1)), mincost[1] = 0; int ans = 0, num = 0; while(! que.empty()) { int v = que.top().second; que.pop(); if(vis[v]) continue; vis[v] = true; num++; ans += mincost[v]; for(int i = head[v]; i != -1; i = g[i].next) { int u = g[i].to; if(mincost[u] > g[i].cost) { mincost[u] = g[i].cost; que.push(pii(mincost[u], u)); } } } return num == n ? ans : -1; } void table() { prime[0] = prime[1] = 1; for(int i = 2; i * i < M; i++) if(! prime[i]) for(int j = i + i; j < M; j += i) prime[j] = 1; } int main() { table(); int t, n, a[N]; scanf("%d", &t); while(t--) { cnt = 0; memset(head, -1, sizeof head); scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) { if(! prime[a[i]] || ! prime[a[j]] || ! prime[a[i]+a[j]]) { int cost = min(min(a[i], a[j]), abs(a[i] - a[j])); add_edge(i, j, cost), add_edge(j, i, cost); } } printf("%d\n", prim(n)); } return 0; }